Skip to content
dronacharya.ai
← JEE Advanced 2014 · Paper 2
MathematicsList match+3 / -1Question 60 of 60

JEE Advanced 2014 · Paper 2, Mathematics question 60

Let f1:R→Rf_1:\mathbb R\to\mathbb R, f2:[0,∞)→Rf_2:[0,\infty)\to\mathbb R, f3:R→Rf_3:\mathbb R\to\mathbb R and f4:R→[0,∞)f_4:\mathbb R\to[0,\infty) be defined by
f1(x)={∣x∣if x<0,exif x≥0;f2(x)=x2;f_1(x)=\begin{cases}|x| & \text{if } x<0,\\ e^x & \text{if } x\ge0;\end{cases}\qquad f_2(x)=x^2;
f3(x)={sin⁡xif x<0,xif x≥0f_3(x)=\begin{cases}\sin x & \text{if } x<0,\\ x & \text{if } x\ge0\end{cases}
and
f4(x)={f2(f1(x))if x<0,f2(f1(x))−1if x≥0.f_4(x)=\begin{cases}f_2(f_1(x)) & \text{if } x<0,\\ f_2(f_1(x))-1 & \text{if } x\ge0.\end{cases}

List II

  • P
    f4f_4 is
  • Q
    f3f_3 is
  • R
    f2∘f1f_2\circ f_1 is
  • S
    f2f_2 is

List I

  • 1
    onto but not one-one
  • 2
    neither continuous nor one-one
  • 3
    differentiable but not one-one
  • 4
    continuous and one-one
Choose one option
Answer and worked solution

Official answer: D

The functions. f2∘f1(x)=x2f_2\circ f_1(x)=x^2 for x<0x<0 and e2xe^{2x} for x≥0x\ge0. So
f4(x)={x2x<0e2x−1x≥0.f_4(x)=\begin{cases}x^2&x<0\\e^{2x}-1&x\ge0.\end{cases}
(The paper labels the left list "List II" and the right list "List I"; the code reads the same either way.)

  • P, f4f_4: continuous at 0, range [0,∞)[0,\infty) (onto its codomain). Not one-one: x2x^2 for x<0x<0 repeats values of e2x−1e^{2x}-1. (Its one-sided derivatives at 0 are 0 and 2, so it is not differentiable.) Onto but not one-one → 1.
  • Q, f3f_3: both pieces have derivative 1 at 0, so it is differentiable; sin⁡x\sin x repeats values for x<0x<0. Differentiable but not one-one → 3.
  • R, f2∘f1f_2\circ f_1: jumps from 00 to 11 at x=0x=0, so it is discontinuous; values above 1 are taken twice. Neither continuous nor one-one → 2.
  • S, f2=x2f_2=x^2 on [0,∞)[0,\infty): continuous and one-one → 4.

(D) P → 1, Q → 3, R → 2, S → 4